Unformatted text preview: with n1 degrees of freedom. So, for the desired probability, we can write. Pr(0 . 03 < S < . 07) = Pr(0 . 0009 < S 2 < . 0049) = Pr â€¡ . 0009 Â· 19 . 0025 < X 2 < . 0049 Â· 19 . 0025 Â· = Pr(6 . 84 < X 2 < 37 . 24) = 0 . 995. 007 = 0 . 988 . 4. We know that Ë† P âˆ¼ N Â± p, q p (1p ) n Â¶ . So for the probability in question, we obtain Pr( Ë† P < . 60) = Î¦ . 60. 70 q . 70 Â· . 30 200 = Î¦(3 . 09) = 0 . 001 . 1...
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 Spring '08
 Perevalov
 Degrees Of Freedom, Chisquare distribution, average observed biases, average biases decrease

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